Additional Mathematics 0606 / Calculus — Differentiation 2 / 3.993.99: Tangent to exponential curve0606/22/M/J/24 — Question 5 · 6 marksMark as done · Save for later · Show all solutionsA curve has equation y=5ex+e−2xy = 5\mathrm{e}^{x} + \mathrm{e}^{-2x}y=5ex+e−2x. The tangent at x=1x = 1x=1 cuts the xxx-axis at PPP. Find the equation of the tangent in the form y=mx+cy = mx + cy=mx+c and hence the xxx-coordinate of PPP.[6]▸ Answer▸ Official mark scheme← 3.38: Small change for cos x sin 2x3.10: Small change for a logarithmic product →